Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 10002;
int a[Maxn], b[Maxn];
int loc[32];
int n;
vector<pair<int, int> > op[2];
void getbase(int *x, int t) {
  for (int i = 0, row = 0; i < n && row < 32; i++, row++) {
    int cs = -1;
    for (int j = i; j < n; j++)
      if (x[j] >> row & 1) {
        cs = j;
        break;
      }
    if (cs < 0) {
      i--;
      continue;
    }
    if (cs != i) {
      op[t].push_back(pair<int, int>(i, cs));
      op[t].push_back(pair<int, int>(cs, i));
      op[t].push_back(pair<int, int>(i, cs));
      swap(x[i], x[cs]);
    }
    if (!t) loc[row] = i;
    for (int j = 0; j < n; j++)
      if (j != i && (x[j] >> row & 1)) {
        op[t].push_back(pair<int, int>(j, i));
        x[j] ^= x[i];
      }
  }
}
void pt(int x) {
  for (int i = 0; i < 32; i++) printf("%d", x >> i & 1);
  puts("");
}
bool solve() {
  memset(loc, -1, sizeof(loc));
  getbase(a, 0);
  getbase(b, 1);
  for (int i = 0, row = 0; i < n; i++) {
    int maxst = min(n, 32);
    if (!b[i]) {
      op[0].push_back(pair<int, int>(i, i));
      continue;
    }
    while (!(b[i] >> row & 1)) row++;
    int cs = -1;
    for (int j = i; j < maxst; j++)
      if (!(a[j] >> row) && ((a[j] >> row) > 1))
        op[0].push_back(pair<int, int>(j, j)), a[j] = 0;
      else if (a[j] >> row & 1) {
        cs = j;
        break;
      }
    if (cs < 0) return false;
    if (cs != i) {
      op[0].push_back(pair<int, int>(i, cs));
      op[0].push_back(pair<int, int>(cs, i));
      op[0].push_back(pair<int, int>(i, cs));
      swap(a[i], a[cs]);
    }
    for (int t = row + 1; t < 32; t++) {
      if ((b[i] >> t & 1) != (a[i] >> t & 1)) {
        if (loc[t] < 0) return false;
        a[i] ^= a[loc[t]];
        op[0].push_back(pair<int, int>(i, loc[t]));
      }
    }
    if (a[i] != b[i]) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) scanf("%d", b + i);
  if (!solve())
    puts("-1");
  else {
    printf("%d\n", (int)op[0].size() + (int)op[1].size());
    for (int i = 0; i < op[0].size(); i++)
      printf("%d %d\n", op[0][i].first + 1, op[0][i].second + 1);
    for (int i = op[1].size() - 1; i >= 0; i--)
      printf("%d %d\n", op[1][i].first + 1, op[1][i].second + 1);
  }
}